# Math Help - A few cool ones here

1. ## A few cool ones here

simple. How many distinct ways can the letters in the word: MISSISSIPPI be arranged.

another one:

what is the units digit of 9^4321 power?

another one:

If Bob spends 4 hours to completely repair a roof, and Frank spends 6 hours to completely repair that same roof, how long would it take them to repair the roof together.

another one:

How many terminating zeros are in 4321!

just a little note guys. All of these are from past math competitions i have been in. Lets see if you guys can crack them.

2. Originally Posted by rtblue
simple. How many distinct ways can the letters in the word: MISSISSIPPI be arranged. This comes up in a BUNCH of math competitions in the 7th and 8th grade. Lets see if you guys know how to do it
This is a pretty standard combinatorial problem. We have eleven characters, with four I's, four S's, and 2 P's. So, the total possible permutations are

$\frac{11!}{4!\,4!\,2!}=\frac{11\cdot10\cdot9\cdot8 \cdot7\cdot6\cdot5}{4\cdot3\cdot2^2}$

$=34\,650.$

Edit:
what is the units digit of 9^4321 power?
As you increase $n,$ the units digit of $9^n$ alternates between 1 and 9. Why? Because for arbitrary digits $d_i,$

$d_n\dots d_3d_2d_19\cdot9$

$=\left(10^nd_n+\cdots+10^2d_2+10^1d_1+9\right)\cdo t9$

$=9\cdot10^nd_n+\cdots+9\cdot10^2d_2+9\cdot10^1d_1+ 81,$

which has a units digit of 1, and it can be similarly shown that any integer ending in 1, when multiplied by 9, will give an integer ending in 9 (because $1\cdot9=9$).

So, for even $n,\;9^n$ ends in a 1, and for odd $n,\;9^n$ ends in a 9. We have the odd case, so the units digit is nine.

If Bob spends 4 hours to completely repair a roof, and Frank spends 6 hours to completely repair that same roof, how long would it take them to repair the roof together.
Bob repairs roofs at a rate of $\frac14\text{ roofs}/\text{hr.}$ and Frank repairs at a rate of $\frac16\text{ roofs}/\text{hr.}$ Now, assuming these two slouches don't know the meaning of the word "synergy," their combined repair rate is

$\frac14+\frac16=\frac5{12}\text{ roofs}/\text{hr.}$

Thus, to repair this one roof, Bob and Frank will need to work for

$\left(1\text{ roof}\right)\left(\frac{12\text{ hr.}}{5\text{ roofs}}\right)=\frac{12}5\text{ hr.}$

$=2.4\text{ hr.}=2\text{ hr., }24\text{ min.}$

How many terminating zeros are in 4321!
This is also a pretty standard problem. The number of zeros corresponds to the number of factors of 10, which is the number of pairs of 2 and 5 in the number's factorization. In $4321!,$ every second number in this factorial is even (divisible by two), but only every fifth factor is divisible by five. So, since there are plenty of 2's, the number of factors of 10 is limited only by the factors of 5.

$\frac{4321}5=864.2,$ so there are 864 numbers between 1 and 4321 that have at least one factor of 5. Similarly, $\frac{4321}{5^2}=172.84,$ which means that 172 of these numbers also have a second factor of 5. Continuing in this way, the total of number of factors of 5 is

$\begin{tabular}{crcr}
&\lfloor4321/5\rfloor&=&864\\
&\lfloor4321/5^2\rfloor&=&172\\
&\lfloor4321/5^3\rfloor&=&34\\
&\lfloor4321/5^4\rfloor&=&6\\
+&\lfloor4321/5^5\rfloor&=&1\\\hline
&&&1077
\end{tabular}.$

So, there are 1077 trailing zeros.

3. GREAT, you answered all of the correctly. I must wonder, you don't have to tell me, what grade are you in?